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Question
Find the values of k for which the roots are real and equal in each of the following equation:
kx2 + kx + 1 = -4x2 - x
Solution
The given quadric equation is kx2 + kx + 1 = -4x2 - x, and roots are real and equal
Then find the value of k.
Here,
kx2 + kx + 1 = -4x2 - x
4x2 + kx2 + kx + x + 1 = 0
(4 + k)x2 + (k + 1)x + 1 = 0
So,
a = (4 + k), b = (k + 1) and c = 1
As we know that D = b2 - 4ac
Putting the value of a = (4 + k), b = (k + 1) and c = 1
= (k + 1)2 - 4 x (4 + k) x (1)
= (k2 + 2k + 1) - 16 - 4k
= k2 - 2k - 15
The given equation will have real and equal roots, if D = 0
Thus,
k2 - 2k - 15 = 0
Now factorizing of the above equation
k2 - 2k - 15 = 0
k2 - 5k + 3k - 15 = 0
k(k - 5) + 3(k - 5) = 0
(k - 5)(k + 3) = 0
So, either
k - 5 = 0
k = 5
Or
k + 3 = 0
k = -3
Therefore, the value of k = 5, -3.
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